Question: Let $g(x)=\sec(x)$. Find $g'\left(\dfrac{3\pi}{4}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $1$ (Choice C) C $-\sqrt{2}$ (Choice D) D $\sqrt{2}$
Answer: Let's first find $g'(x)$. Then, we can evaluate it at $x=\dfrac{3\pi}{4}$. Recall that the derivative of $\sec(x)$ is $\dfrac{\sin(x)}{\cos^2(x)}$, or $\sec(x)\tan(x)$. [Is there a way to know this without memorizing?] So $g'(x)=\dfrac{\sin(x)}{\cos^2(x)}$. Now let's plug $x={\dfrac{3\pi}{4}}$ into $g'$ : $\begin{aligned} &\phantom{=}g'\left({\dfrac{3\pi}{4}}\right) \\\\ &=\dfrac{\sin\left({\dfrac{3\pi}{4}}\right)}{\cos^2\left({\dfrac{3\pi}{4}}\right)} \\\\ &=\dfrac{\dfrac{\sqrt2}{2}}{\left(-\dfrac{\sqrt2}{2}\right)^2} \\\\ &={\dfrac{\sqrt2}{2}}\cdot{\dfrac{4}{2}} \\\\ &=\sqrt{2} \end{aligned}$ In conclusion, $g'\left(\dfrac{3\pi}{4}\right)=\sqrt{2}$.